What is the answer to log_{2}((k+2)/(3-k))=-2 ?

The answer to log_{2}((k+2)/(3-k))=-2 is k=-1

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log_{2}((k+2)/(3-k))=-2

lo g_{2} (\frac{k + 2}{3 - k}) = - 2

k = - 1

Solution steps

lo g_{2} (\frac{k + 2}{3 - k}) = - 2

Apply log rules

\frac{k + 2}{3 - k} = 2^{- 2}

Multiply both sides by

3 - k

\frac{k + 2}{3 - k} (3 - k) = 2^{- 2} (3 - k)

Simplify

2^{- 2} (3 - k) : \frac{1}{4} (3 - k)

k + 2 = \frac{1}{4} (3 - k)

Solve

k + 2 = \frac{1}{4} (3 - k) : k = - 1

k = - 1

Verify Solutions:

k = - 1

True

The solution is

k = - 1

2.8 = - lo g_{10} (x)

lo g_{3} (x + 7) + lo g_{3} (x + 5) = 1

lo g_{b} (8) = - \frac{1}{2}

1 + ln (3 x) = 0

lo g_{10} (x) = 0.17527

What is the answer to log_{2}((k+2)/(3-k))=-2 ?
The answer to log_{2}((k+2)/(3-k))=-2 is k=-1
Solve for k: log_{2}((k+2)/(3-k))=-2
The solution is k=-1